JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 4)
Let $$\mathrm{g}(x)=f(x)+f(1-x)$$ and $$f^{\prime \prime}(x) > 0, x \in(0,1)$$. If $$\mathrm{g}$$ is decreasing in the interval $$(0, a)$$ and increasing in the interval $$(\alpha, 1)$$, then $$\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$$ is equal to :
$$\frac{3 \pi}{4}$$
$$\pi$$
$$\frac{5 \pi}{4}$$
$$\frac{3 \pi}{2}$$
Explanation
We have, $g(x)=f(x)+f(1-x)$
Differentiating both side, we get
$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$
As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.
Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$
So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$
$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.
Now,
$$ \begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned} $$
Differentiating both side, we get
$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$
As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.
Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$
So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$
$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.
Now,
$$ \begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned} $$
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